# Number of distinct scheme structures on a set [closed]

Given a cardinal number $$|X|$$, how many isomorphism classes of schemes with the cardinality of the set of points equal to $$|X|$$ are there?

## closed as off-topic by Steven Landsburg, Dan Petersen, user44191, abx, Todd Trimble♦Jun 18 at 16:32

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• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Steven Landsburg, Dan Petersen, user44191, abx, Todd Trimble
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For any cardinal $$\kappa\neq 0$$, there is a proper class of schemes of cardinality $$\kappa$$ up to isomorphism. (In the case $$\kappa = 0$$, there is a unique empty scheme up to isomorphism, namely the spectrum of the zero ring.)
Note first that there is a proper class of schemes with just $$1$$ point. Indeed, there are fields of every infinite cardinality, and for any field $$k$$, $$\text{Spec}(k)$$ has only one point. This observation easily extends to the case when $$\kappa$$ is nonzero and finite, since for any field $$k$$, $$\text{Spec}(k^n)$$ has $$n$$ points.
In the case when $$\kappa$$ is infinite, we cannot handle arbitrary $$\kappa$$ with an infinite product of fields. For example, the points of $$\text{Spec}(k^\mathbb{N})$$ are in bijection with the ultrafilters on $$\mathbb{N}$$, so $$|\text{Spec}(k^\mathbb{N})| = 2^{2^{\aleph_0}}$$.
Instead, we can use the fact that $$\kappa = \kappa+1$$ when $$\kappa$$ is infinite. Fix an algebraically closed field $$F$$ of cardinality $$\kappa$$. For any field $$k$$, we have $$|\text{Spec}(F[x]\times k)| = |\text{Spec}(F[x])\sqcup \text{Spec}(k)| = \kappa$$, since $$\text{Spec}(F[x])$$ has one point for every element of $$F$$, together with a single generic point, and $$\text{Spec}(k)$$ has a single point. So again we have a proper class of schemes with underlying set of cardinality $$\kappa$$.
• You could also just take the disjoint union of $\kappa$ copies of $\mathrm{Spec}\,k$ (of course your construction is nicer because it produces an affine scheme) – Denis Nardin Jun 18 at 14:05